Problem: $\sum\limits_{n=1}^{\infty } \dfrac{x^n}{n\cdot5^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $0 \le x \le 10$ (Choice B) B $0 \le x < 10$ (Choice C) C $-5 \le x \le 5$ (Choice D) D $-5 \le x<5$
Explanation: We use the ratio test. For $x\neq 0$, let $a_n= \dfrac{x^n}{n\cdot5^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x}{5}\right| $ The series converges when $\left| \dfrac{x}{5}\right|<1$, which is when $-5<x<5$. Now let's check the endpoints, $x=-5$ and $x=5$. Letting $x=-5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-5)^n}{n\cdot5^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot 5^n}{n\cdot5^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n} \end{aligned}$ This is the alternating harmonic series, which is known to converge. Letting $x=5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{5^n}{n\cdot5^n} &=\sum\limits_{n=1}^{\infty } \dfrac{1}{n} \end{aligned}$ This is the harmonic series, which is known to diverge. In conclusion, the interval of convergence is $-5 \le x<5$.